What is the lim (x→0) (cosmx-cosnx)/x^2 ?

Question

my answer is (-m^2+n^2)/2

can anyone verify this?

Answer 1

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Answer 2

lim x-----> 0 (cos mx - cos nx)/x^2

= lim x------->0  [- 2sin (m+n)x/2 .sin (m-n)x/2]/x^2

= - 2[lim x---->0 sin((m+n)x/2)/x]  x [lim x---->0 sin((m-n)x/2)/x]

                   ...............(1)

take, (m+n)x/2 = z => when x----> 0 => z ----> 0

and 2z/(m+n)

similarly, take, (m -n)x/2 = p

so, when x--->0 => p--->0

and x = 2p/(m -n)

now,

lim x----> 0 sin((m+n)x/2)/x

= lim z---->0 sin z/(2z/(m+n))

= lim z---->0 (m+n)/2 . sin z/z

= (m+n)/2 . lim z-->0 sin z/z

=(m+n)/2 [since, lim z---> 0 sin z/z = 1]

similarly,

lim x----> 0 sin((m - n)x/2)/x

= lim p-----> 0 sin p/(2p/(m-n))

= (m - n)/2 lim p ---> 0 sin p/p

= (m - n)/2

so now we have,

lim x --->0 (cos mx - cos nx)/x^2

= - 2[(m+n)/2][(m - n)/2]

= - 2(m^2 - n^2)/4

= - (m^2 - n^2)/2

= (- m^2 + n^2)/2 <==ANSWER

Hope I helped u :)

Answer 3

If you haven't learned L'Hopital's rule, you can still get it by using trig. identity.

cosmx - cosnx

= cos[(m+n)x/2 + (m-n)x/2] - cos[(m+n)x/2 - (m-n)x/2]

= -2sin[(m+n)x/2]sin[(m-n)x/2]

So,

lim{x->0}(cosmx-cosnx)/x^2

= lim{x->0} -2sin[(m+n)x/2]sin[(m-n)x/2]/x^2

= (-1/2)(m+n)(m-n)

= (-1/2)(m^2 - n^2)

= (1/2)(n^2 -m^2)

By the way, the answer covers the case m = n.

Answer 4

0/0 so use L'Hopitals

lim (x→0) (cos(mx) - cos(nx))/x²

= lim (x→0) (- msin(mx) + nsin(nx))/2x

= lim (x→0) (nsin(nx) - msin(mx)) /2x

still 0 / 0 so do it again

= lim (x→0) (n²cos(nx) - m²cos(mx))/2

= (n² - m²)/2

so yes you are correct

Answer 5

Case 1 : m = n

=> lim ((x → 0) (cosmx - cosnx)/x^2

= 0

(because cosmx - cosnx = cosmx - cosmx = 0)

Case 2 : m ≠ n

=> lim ((x → 0) (cosmx - cosnx)/x^2

= lim [(x → 0) (2sin(m+n)x/2*sin(n-m)x/2] / x^2

= (n^2 - m^2)/2 lim [(x → 0) [sin(m+n)x/2 * sin(n-m)x/2] / (n^2 - m^2)x^2/4

= (n^2 - m^2) / 2

=> your answer is correct only if m ≠ n.

For m = n, the value of the limit is 0.